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laiz [17]
4 years ago
5

A cop claims you passed him at more than 60 mph ,27.8 meters per second, 200 meters from a red light that just turned green. you

drive a car that accelerates at a max of 1.5 meters per second. Is the cop correct?
Physics
1 answer:
postnew [5]4 years ago
5 0

Sounds like the situation is modeled like so:

[Light] - - - - - - - - - - [You] => - - -

- - - - - - - - - - - - - - - [Cop] - - - - -

<= = = 200 m = = =>

where your velocity is 27.8 m/s to the right. So the question is asking, is it possible that your car, given that it accelerates at a maximum of 1/5 m/s^2, can reach a speed of 27.8 m/s from rest in the time it takes for it to cover 200 m.

The car's position at time t is

x=\dfrac12\left(1.5\dfrac{\rm m}{\mathrm s^2}\right)t^2

The time it takes to traverse 200 m is

200\,\mathrm m=\left(0.75\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t\approx16.3\,\mathrm s

Your car's velocity at time t, starting from rest, is

v=\left(1.5\dfrac{\rm m}{\mathrm s^2}\right)t

so that after 16.3 s, your car is moving at a speed of

v=\left(1.5\dfrac{\rm m}{\mathrm s^2}\right)\left(16.3\,\mathrm s\right)\approx24.5\dfrac{\rm m}{\rm s}

which is less than the speed the cop claims you were going, so the cop is not correct.

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The answer is B. X-rays and gamma rays. The electromagnetic spectrum consists of all types of electromagnetic radiation. The EM spectrum is arranged from the longest wavelength with the lowest frequency and lowest energy which are the radio waves and microwaves to the shortest wavelength with the highest frequency and highest energy which are the x-rays and gamma rays. Visible light that can be seen by a humans is a small portion in between the ultraviolet and infrared rays.
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A truck is speeding up as it travels on an interstate. The truck's momentum (in kg · m/s) is proportional to the truck's speed (
serious [3.7K]

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a)48300 times

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Explanation:

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So we can say that truck momentum is 48300 times its velocity.

Now given that

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P = 48300 x 17  kg · m/s

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6 0
4 years ago
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A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1
baherus [9]

Answer:

The  value of the power is   P_c  =  38.55 \  W

Explanation:

From the question we are told that

   The  power  rating P_{1000} =P_b=  52 \  W

    The frequency is  f = 1000 \  Hz

    The  frequency at which the sound intensity decreases  f_k  =  20 \  Hz

     The decrease in intensity is by \beta  =  1.3 dB

Generally the  initial intensity of the speaker  is mathematically represented as

     \beta_1 =  10 log_{10} [\frac{P_b}{P_a} ]

Generally the intensity of the speaker after it has been decreased is

       \beta_2 =  10 log_{10} [\frac{P_c}{P_a} ]

So

\beta_1-\beta_2 =  10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]

=>  \beta =  10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3

=>  \beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3

=>  \beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3

=> 10log_{10} [\frac{P_b}{P_c} ] = 1.3

=> log_{10} [\frac{P_b}{P_c} ] = 0.13

taking atilog of both sides

[\frac{P_b}{P_c} ] = 10^{0.13}      

=>[\frac{52}{P_c} ] = 10^{0.13}      

=>  P_c  =  \frac{52}{1.34896}

=>   P_c  =  38.55 \  W

   

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